First, consider 3 sequences $\left(x_{n}\right)$, $\left(y_{n}\right)$, $\left(z_{n}\right)$, $n \geq 1$ with
$$x_{1} = |x - y|, \quad y_{1} = |y - z|, \quad z_{1} = |z - x| \text{ and } \left\{ \begin{array}{l} x_{n+1} = \left|x_{n} - y_{n}\right| \\ y_{n+1} = \left|y_{n} - z_{n}\right| \\ z_{n+1} = \left|z_{n} - x_{n}\right| \end{array} \right.$$
From this, it is easy to see that $x_{n}$, $y_{n}$, $z_{n} \geq 0$ for all $n$.
Let $w_{n} = \max \left\{x_{n}, y_{n}, z_{n}\right\}$, $n \geq 1$. We have
$$\begin{aligned} w_{n+1} = \max \left\{x_{n+1}, y_{n+1}, z_{n+1}\right\} & = \max \left\{\left|x_{n} - y_{n}\right|, \left|y_{n} - z_{n}\right|, \left|z_{n} - x_{n}\right|\right\} \\ & \leq \max \left\{x_{n}, y_{n}, z_{n}\right\} = w_{n}. \end{aligned}$$
The equality occurs when there is at least one number among $x_{n}$, $y_{n}$, $z_{n}$ equal to $0$. Hence, the sequence $(w_{n})$ is non-increasing. Suppose that $k$ is a positive integer such that $(x_{k}, y_{k}, z_{k}) = (x, y, z)$, then
$$w_{1} \geq w_{2} \geq \ldots \geq w_{k} = w_{1} \text{ which implies that } w_{1} = w_{2} = \ldots = w_{k}.$$
On the other hand, for all $i = \overline{1, k}$, there is at least one number among $(x_{i}, y_{i}, z_{i})$ equal to $0$. Without loss of generality, we may assume that $x \geq y \geq z = 0$ and we can see that $x_{1} = |x - y|$, $y_{1} = |y - z| = y$, $z_{1} = |z - x| = x$ so we have some cases:

1. If $|x - y| = 0 \Leftrightarrow x = y$, we have the tuple $(x, y, z) = (a, a, 0)$ with $a \geq 0$. It is easy to check this tuple satisfies the given condition.

2. If $y = 0$, to separate from the previous case, we consider $x > 0$, we have the tuple $(x, y, z) = (a, 0, 0)$ with $a > 0$. But $x_{1} = a$, $y_{1} = 0$, $z_{1} = a$ and from this, we cannot obtain $(a, 0, 0)$ anymore. Hence, this case does not satisfy the given condition.

3. If $x = 0$, we have $x = y = z = 0$ which was mentioned above.

Therefore, the tuples $(x, y, z) = (a, a, 0)$ and its permutations with $a \geq 0$ satisfy the given condition. $\square$